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            <h1 id="seo-header">『算法-ACM竞赛-图论』2-SAT-HDOJHDU1824Let&#39;sgohome</h1>
            
            
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                <h1 id="『算法-ACM-竞赛-图论』2-SAT-HDOJHDU1824Let’sgohome"><a href="#『算法-ACM-竞赛-图论』2-SAT-HDOJHDU1824Let’sgohome" class="headerlink" title="『算法-ACM 竞赛-图论』2-SAT-HDOJHDU1824Let’sgohome"></a>『算法-ACM 竞赛-图论』2-SAT-HDOJHDU1824Let’sgohome</h1><h1 id="图论–2-SAT–HDOJ-HDU-1824-Let’s-go-home"><a href="#图论–2-SAT–HDOJ-HDU-1824-Let’s-go-home" class="headerlink" title="图论–2-SAT–HDOJ&#x2F;HDU 1824 Let’s go home"></a>图论–2-SAT–HDOJ&#x2F;HDU 1824 Let’s go home</h1><p>Problem Description<br>小时候，乡愁是一枚小小的邮票，我在这头，母亲在那头。<br>—— 余光中</p>
<p>集训是辛苦的，道路是坎坷的，休息还是必须的。经过一段时间的训练，lcy 决定让大家回家放松一下，但是训练还是得照常进行，lcy 想出了如下回家规定，每一个队（三人一队）或者队长留下或者其余两名队员同时留下；每一对队员，如果队员 A 留下，则队员 B 必须回家休息下，或者 B 留下，A 回家。由于今年集训队人数突破往年同期最高记录，管理难度相当大，lcy 也不知道自己的决定是否可行，所以这个难题就交给你了，呵呵，好处嘛~，免费**漂流一日。</p>
<p>Input<br>第一行有两个整数，T 和 M，1&lt;&#x3D;T&lt;&#x3D;1000 表示队伍数，1&lt;&#x3D;M&lt;&#x3D;5000 表示对数。<br>接下来有 T 行，每行三个整数，表示一个队的队员编号，第一个队员就是该队队长。<br>然后有 M 行，每行两个整数，表示一对队员的编号。<br>每个队员只属于一个队。队员编号从 0 开始。</p>
<p>Output<br>可行输出 yes，否则输出 no，以 EOF 为结束。</p>
<p>Sample Input<br>1 2 0 1 2 0 1 1 2 2 4 0 1 2 3 4 5 0 3 0 4 1 3 1 4</p>
<p>Sample Output<br>yes no</p>
<p>（1）题目第一个条件：每一个队或者队长留下或者其与两名队员同时留下，或者表明只能为两种情况中的一种；假设三人为 A,B,C，队长为 A，0 表示不留下，1 表示留下，因为 B 与 C 同时留下或者不留下，只要 B，C 中其中一个没有留下或者留下，则 B,C 中另一个也同样留下或者不留下，所以可以从该条件中推导出六条等价关系，即 A 不留下-&gt;B，C 同时留下，A 留下-&gt;B，C 同时不留下，B 留下-&gt;C 留下，A 不留下，B 留下-&gt;C 留下，A 不留下，C 留下-&gt;B 留下，A 不留西，C 不留下-&gt;B 不留下，A 留下；</p>
<p>（2）题目中第二个条件：每一对队员，如果队员 A 留下，则 B 必须回家休息，或者 B 留下，A 必须回家休息；则可以推导出两条等价式：A 留下-&gt;B 不留下，B 留下-&gt;A 不留下，注意在这个条件中可以 A，B 都不留下；</p>
<p>AC 代码：</p>
<pre><code class="hljs">#include &lt;cstdio&gt;
#include &lt;cstring&gt;
#include &lt;queue&gt;
#include &lt;vector&gt;
#include &lt;stack&gt;
#include &lt;algorithm&gt;
#include &lt;iostream&gt;
#define MAXN 7000+10
#define MAXM 400000
#define INF 1000000
using namespace std;
vector&lt;int&gt; G[MAXN];
int low[MAXN], dfn[MAXN];
int dfs_clock;
int sccno[MAXN], scc_cnt;
stack&lt;int&gt; S;
bool Instack[MAXN];
int N, M;
void init()
&#123;
    for(int i = 0; i &lt;=6*N; i++) G[i].clear();
    while(S.size()) S.pop();
    dfs_clock=scc_cnt=0;
&#125;
void getMap()
&#123;
    int a,b,c;
    for(int i=0;i&lt;N;i++)
    &#123;
       scanf(&quot;%d %d %d&quot;,&amp;a,&amp;b,&amp;c);
        G[a+N*3].push_back(b);
        G[a+N*3].push_back(c);
        G[b+3*N].push_back(a);
        G[c+3*N].push_back(a);
    &#125;
    while(M--)
    &#123;
        scanf(&quot;%d%d&quot;,&amp;a,&amp;b);
        G[a].push_back(b+N*3);
        G[b].push_back(a+N*3);
    &#125;

&#125;
void tarjan(int u, int fa)
&#123;
    int v;
    low[u] = dfn[u] = ++dfs_clock;
    S.push(u);
    Instack[u] = true;
    for(int i = 0; i &lt; G[u].size(); i++)
    &#123;
        v = G[u][i];
        if(!dfn[v])
        &#123;
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        &#125;
        else if(Instack[v])
        low[u] = min(low[u], dfn[v]);
    &#125;
    if(low[u] == dfn[u])
    &#123;
        scc_cnt++;
        for(;;)
        &#123;
            v = S.top(); S.pop();
            Instack[v] = false;
            sccno[v] = scc_cnt;
            if(v == u) break;
        &#125;
    &#125;
&#125;
void find_cut(int l, int r)
&#123;
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(sccno, 0, sizeof(sccno));
    memset(Instack, false, sizeof(Instack));
    dfs_clock = scc_cnt = 0;
    for(int i = l; i &lt;= r; i++)
    if(!dfn[i]) tarjan(i, -1);
&#125;
void solve()
&#123;
    for(int i = 0; i &lt; 3*N; i++)
    &#123;
        if(sccno[i] == sccno[i+3*N])//矛盾
        &#123;
            printf(&quot;no\n&quot;);
            return ;
        &#125;
    &#125;
    printf(&quot;yes\n&quot;);
&#125;
int main()
&#123;
    while(scanf(&quot;%d%d&quot;, &amp;N, &amp;M) != EOF)
    &#123;
        init();
        getMap();
        find_cut(0, 6*N-1);
        solve();
    &#125;
    return 0;
&#125;
</code></pre>

                
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